Solution :
Capacitor of the capacitance, $C = 12\; pF = 12 \times 10^{−12} F$
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
$e= \large\frac{1}{2} $$CV^2 =\frac{1}{2} \times 12 \times 10^{-12} \times (50)^2J = 1.5 \times 10^{-8}\;J$
Therefore, the electrostatic energy stored in the capacitor is $1.5 \times 10^{-8}\; J$